In fig. 8.23, ∆PQR and ∆SQR are isosceles.
i) find ∆PQR and ∆SRQ
ii) find ∆SQR and ∆SRQ
iii) find x
Answers
Hi there,
I think the information given in the question is incorrect. Hoping the correct question to be as follows:
In fig. 8.23, ∆PQR and ∆SQR are isosceles.
i) find angle PQR and angle PRQ
ii) find angle SQR and ∆SRQ
iii) find x
Answer:
i) ∠ PQR = ∠PRQ = 75°
ii)∠SQR = ∠SRQ = 55°
iii) ∠x = 20°
Step-by-step explanation:
Case 1:
We are given ∆ PQR is an isosceles triangle, i.e.,
PQ = PR
∴ ∠PQR = ∠PRQ [∵ angles opposite to equal sides are equal] …… (i)
According to the angle sum property, we have
∠P + ∠PQR + ∠PRQ = 180°
⇒30° + 2 (∠PQR) = 180° …. [∠P = 30° given]
⇒ 2 (∠PQR) = 180° - 30° = 150° …. [from (i)]
⇒ ∠PQR = 150° / 2 = 75°
∴ ∠ PQR = ∠PRQ = 75° ….. (ii)
Case 2:
Similarly, we are given ∆ SQR is an isosceles triangle, i.e.,
SQ = SR
∴ ∠SQR = ∠SRQ [∵ angles opposite to equal sides are equal] …… (iii)
According to the angle sum property, we have
∠S + ∠SQR + ∠SRQ = 180°
⇒70° + 2 (angle SQR) = 180° …. [angle P = 70° given]
⇒ 2 (angle SQR) = 180° - 70° = 110° …. [from (iii)]
⇒ angle SQR = 110° / 2 = 55°
∴ ∠SQR = ∠SRQ = 55° ….. (iv)
Case 3:
Subtracting (ii) from (i), we get
∠x
= ∠PQS
= ∠PQR - ∠SQR
= 75° - 55°
= 20°
Answer:
i) ∠ PQR = ∠PRQ = 75°
ii)∠SQR = ∠SRQ = 55°
iii) ∠x = 20°
Step-by-step explanation:
Case 1:
We are given ∆ PQR is an isosceles triangle, i.e.,
PQ = PR
∴ ∠PQR = ∠PRQ [∵ angles opposite to equal sides are equal] …… (i)
According to the angle sum property, we have
∠P + ∠PQR + ∠PRQ = 180°
⇒30° + 2 (∠PQR) = 180° …. [∠P = 30° given]
⇒ 2 (∠PQR) = 180° - 30° = 150° …. [from (i)]
⇒ ∠PQR = 150° / 2 = 75°
∴ ∠ PQR = ∠PRQ = 75° ….. (ii)
Case 2:
Similarly, we are given ∆ SQR is an isosceles triangle, i.e.,
SQ = SR
∴ ∠SQR = ∠SRQ [∵ angles opposite to equal sides are equal] …… (iii)
According to the angle sum property, we have
∠S + ∠SQR + ∠SRQ = 180°
⇒70° + 2 (angle SQR) = 180° …. [angle P = 70° given]
⇒ 2 (angle SQR) = 180° - 70° = 110° …. [from (iii)]
⇒ angle SQR = 110° / 2 = 55°
∴ ∠SQR = ∠SRQ = 55° ….. (iv)
Case 3:
Subtracting (ii) from (i), we get
∠x
= ∠PQS
= ∠PQR - ∠SQR
= 75° - 55°
= 20°
Step-by-step explanation: